Problem: The polynomial $p(x)=5x^3-44x^2+61x+14$ has a known factor of $(x-7)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
We know $(x-7)$ is a factor of $p(x)$. This means that $p(x)=(x-7)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-7)$ $\begin{array}{r} 5x^2-\phantom{1}9x-\phantom{1}2 \\ x-7|\overline{5x^3-44x^2+61x+14} \\ \mathllap{-(}\underline{5x^3-35x^2\phantom{+61x+14}\rlap )} \\ -9x^2+61x+14 \\ \mathllap{-(}\underline{-9x^2+63x\phantom{+14}\rlap )} \\ -2x+14 \\ \mathllap{-(}\underline{-2x+14\rlap )} \\ 0 \end{array}$ We find that $q(x)=5x^2-9x-2$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=5x^2-9x-2 \\\\ &=5x^2-10x+x-2 \\\\ &=5x(x-2)+1(x-2) \\\\ &=(5x+1)(x-2) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=5x^3-44x^2+61x+14 \\\\ &=(x-7)(5x^2-9x-2) \\\\ &=(x-7)(5x+1)(x-2) \end{aligned}$